If the roots of the quadratic equation (a-b) x2 - (b-c) x + (c-a) =0 are equal, prove that 2a=b +c
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The question must be, if roots of equation (a-b)x2+(b-c)x+(c-a)=0 are equal , prove that b+c = 2a.
Using Discriminant,
D = B2-4AC as compared with the general quadratic equation Ax2+Bx+C=0
so, A = a-b
B = b-c
C = c-a
For roots to be equal, D=0
(b-c)2 - 4(a-b)(c-a) =0
b2+c2-2bc -4(ac-a2-bc+ab) =0
b2+c2-2bc -4ac+4a2+4bc-4ab=0
4a2+b2+c2+2bc-4ab-4ac=0
(2a-b-c)2=0
i.e. 2a-b-c =0
2a= b+c
The question must be, if roots of equation (a-b)x2+(b-c)x+(c-a)=0 are equal , prove that b+c = 2a.
Using Discriminant,
D = B2-4AC as compared with the general quadratic equation Ax2+Bx+C=0
so, A = a-b
B = b-c
C = c-a
For roots to be equal, D=0
(b-c)2 - 4(a-b)(c-a) =0
b2+c2-2bc -4(ac-a2-bc+ab) =0
b2+c2-2bc -4ac+4a2+4bc-4ab=0
4a2+b2+c2+2bc-4ab-4ac=0
(2a-b-c)2=0
i.e. 2a-b-c =0
2a= b+c
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